# Fraction Differences

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Introduction

Fraction Differences First Sequence To begin with I looked at the first sequence of fractions to discover the formula that explained it. As all the numerators were 1 I looked at the denominators. As these all increased by 1 every time, I figured that the formula was simply as the denominators corresponded to the implied first line as shown in this table below: nth number 1 2 3 4 5 6 7 8 Denominators 1 2 3 4 5 6 7 8 I shall call this Formula 1 (F1) for easy reference. Second Sequence Again I decided to discount the numerator as it was 1, and I decided to concentrate on the differences between the denominators rather than the 'fractions'. So I am looking for a formula that will explain the sequence: 2, 6, 12, 20, 30. First of all though I decided to extend the sequence in order to have a broader range to work with. I used a calculator to work out the following denominators finding the difference between and , and all the way up to I set the differences out in a table to try to find the pattern: nth number 1 2 3 4 5 6 7 8 9 Sequence 2 6 12 20 30 42 ...read more.

Middle

I also believed that the formula would be in some way connected with one of the previous formulas and decided to test this by working from the same formula as the previous one: Again I looked at the difference between the numbers in the sequence and n�: Sequence n� Difference 3 1 + 2 12 8 + 4 30 27 + 3 60 64 + 4 105 125 + 20 There was no decisive pattern, the coefficient was neither a constant number or an 'n' term. I decided to look at the second correct formula I had deduced for the second sequence n(n + 1) (F2) and thought about how I could extend this to include the n� I knew was necessary for the formula. n(n + 1) (n + 2) I started with the number 1: 1(1 + 1) (1 + 2) = 1(2)(3) = 1 x 2 x 3 = 6 This was twice what I needed (3 for the first number in the sequence) so I divided it by 2. So the formula I was going to attempt was: First attempt, I tried substituting n with 2: The answer was the next number in the sequence, I continued this time replacing n with 3: The formula had worked again, ...read more.

Conclusion

as the denominator. F6 = = 5 = 30 It worked. Beginning Again Now that I had worked out formulas and found the pattern that they too were 'sequencing' by I decided to look at a different set of fractions to see if the patterns applied to them too. Whereas the first sequence had the denominators: 1, 2, 3, 4, 5... I decided to try denominators that went up by 2: 2, 4, 6, 8, 10... I began by setting out the numbers in a table to find the differences: nth number 1 2 3 4 5 6 7 8 Sequence 2 4 6 8 10 12 14 16 First Difference 2 2 2 2 2 2 2 As the first difference was a constant I knew that the formula would contain this difference as the coefficient of n: 2n. I now tried a sequence where the numbers went up by 3: nth number 1 2 3 4 5 6 7 8 Sequence 3 6 9 12 15 18 21 24 First Difference 3 3 3 3 3 3 3 I saw that the formula would be 3n, 3 the constant first difference would be the coefficient of n. Another pattern had been found. In conclusion I would say that sequences can often throw up many surprises. ?? ?? ?? ?? 6 1 ...read more.

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